Csdrhrt

I found this limit in a book, without any explanation:

$$lim_{ntoinfty}left(sum_{k=0}^{n-1}(zeta(2)-H_{k,2})-H_nright)=1$$

where $H_{k,2}:=sum_{j=1}^kfrac1{j^2}$. However Im unable to find the value of this limit from myself. After some work I get the equivalent expression

$$lim_{ntoinfty}sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$

but anyway Im stuck here. Can someone show me a way to compute this limit? Thank you.

UPDATE: Wolfram Mathematica computed it value perfectly, so I guess there is some integral or algebraic identity from where to calculate it.

Let's see:

$$begin{eqnarray*} sum_{k=0}^{n-1}left(zeta(2)-H_k^{(2)}right) &=& zeta(2)+sum_{k=1}^{n-1}left(zeta(2)-H_k^{(2)}right)\&stackrel{text{SBP}}{=}&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+sum_{k=1}^{n-2}frac{k}{(k+1)^2}\&=&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+(H_{n-1}-1)-sum_{k=1}^{n-2}frac{1}{(k+1)^2}\&=&n( zeta(2)-H_{n-1}^{(2)})+H_{n-1}end{eqnarray*}$$
hence the claim is equivalent to

$$ lim_{nto +infty} n(zeta(2)-H_{n-1}^{(2)}) = lim_{nto +infty}nsum_{mgeq n}frac{1}{m^2} = 1 $$
which is pretty clear since $sum_{mgeq n}frac{1}{m^2} = Oleft(frac{1}{n^2}right)+int_{n}^{+infty}frac{dx}{x^2}=frac{1}{n}+Oleft(frac{1}{n^2}right)$.
$text{SBP}$ stands for Summation By Parts, of course.

Considering your last expression $$a_n=sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$
$$sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}=psi ^{(1)}(k+1)-frac{1}{k+1}$$ making
$$a_n=n ,psi ^{(1)}(n+1)$$ the expansion of which being
$$a_n=1-frac{1}{2 n}+frac{1}{6 n^2}+Oleft(frac{1}{n^4}right)$$